Question Number 37816 by prof Abdo imad last updated on 17/Jun/18
$${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\frac{{ln}\left({x}+{e}^{{sinx}} \right)\:−{x}^{\mathrm{2}} }{{sh}\left(\mathrm{2}{x}\right)} \\ $$
Commented by math khazana by abdo last updated on 28/Jun/18
$${let}\:{use}\:{hospital}\:{theorem}\:{let}\:{f}\left({x}\right)={ln}\left({x}+{e}^{{sinx}} \right)−{x}^{\mathrm{2}} \\ $$$${and}\:{g}\left({x}\right)={sh}\left(\mathrm{2}{x}\right)\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)=\frac{\mathrm{1}+{cosxe}^{{sinx}} }{{x}+{e}^{{sinx}} }\:−\mathrm{2}{x}\:\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} {f}\left({x}\right)=\mathrm{2} \\ $$$${g}^{'} \left({x}\right)=\mathrm{2}{ch}\left(\mathrm{2}{x}\right)\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} {g}^{'} \left({x}\right)=\mathrm{2}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\:\frac{{ln}\left({x}+{e}^{{sinx}} \right)−{x}^{\mathrm{2}} }{{sh}\left(\mathrm{2}{x}\right)}\:=\mathrm{1} \\ $$